Question

1+sinA/cosecA-cotA-1-sinA/cosecA+cotA=2(1+cosA)

Answer 1

Answer:

Step-by-step explanation:

Solving LHS,

1+sin A / cosec A - cot A - 1- sin A/ cosec A+cot A

(1+sin A)(cosec A+ cot A) - (1-sin A)(cosec A- cot A) / cosec 2 A - cot2 A ......[(a+b)(a-b) =a2- b2]

cosec A + cot A + 1 + cos A - cosec A + cot A + 1 - cos A

2 + 2 cot A

2(1+ cot A) = RHS

Answer 2

given: (1+ sinA)/(cosecA- cotA) - (1 - sinA)/(cosecA + cotA) = 2(1+cotA)

to find: prove the above equation.

now to solve such problems, we need to find first LHS and equate it with RHS.

Solving LHS first,

   (1 + sinA/cosecA - cotA) - (1 - sinA/cosecA + cotA)

after converting all trigonometric ratios in terms of sin and cos, we get

   (1+sinA)sinA / (1-cosA) - (1-sinA)sinA / (1+cosA)

Then we need to do cross multiplication,

   {(1+sinA)(1+cosA)sinA - (1-sinA)(1-cosA)sinA}/ (1-cos²A)

Now 1-cos²A is sin²A, So replacing it

   {(1+sinA)(1+cosA)sinA - (1-sinA)(1-cosA)sinA}/ (sin²A)

cancelling sinA from numerator and denominator, and multiply further,

   {1+ sinA + cosA+ sinAcosA - (1- sinA-cosA+sinAcosA)}/ sinA

opening all brackets,

   {1+ sinA + cosA+ sinAcosA - 1+ sinA+cosA-sinAcosA)}/ sinA

we get,

   2sinA+2cosA/sinA

   2(sinA+cosA)/sinA

   2{1+ cotA}      RHS